Problem: The lifespans of bears in a particular zoo are normally distributed. The average bear lives $44.8$ years; the standard deviation is $7.8$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a bear living longer than $60.4$ years.
Answer: $44.8$ $37$ $52.6$ $29.2$ $60.4$ $21.4$ $68.2$ $95\%$ $2.5\%$ $2.5\%$ We know the lifespans are normally distributed with an average lifespan of $44.8$ years. We know the standard deviation is $7.8$ years, so one standard deviation below the mean is $37$ years and one standard deviation above the mean is $52.6$ years. Two standard deviations below the mean is $29.2$ years and two standard deviations above the mean is $60.4$ years. Three standard deviations below the mean is $21.4$ years and three standard deviations above the mean is $68.2$ years. We are interested in the probability of a bear living longer than $60.4$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $95\%$ of the bears will have lifespans within 2 standard deviations of the average lifespan. The remaining $5\%$ of the bears will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({2.5\%})$ will live less than $29.2$ years and the other half $({2.5\%})$ will live longer than $60.4$ years. The probability of a particular bear living longer than $60.4$ years is ${2.5\%}$.